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Solution B15 D2. Conditions of the problem: Load mass m 1 is mounted on a spring suspension in the elevator (Fig. D2.0 - D2.9, Table. D2). Elevator moves vertically by law z = 0,5α1t2 + α2sin (wt) + α3cos (wt) (z-axis vertically upward; z expressed in meters, t - in seconds). On the force of the load resistance of the medium R = μv, where v - speed of cargo with respect to the elevator. Find the cargo movement in relation to the elevator, ie. E. X = f (t); origin placed at a point where the load is attached to the end of the spring when the spring is not deformed. In order to avoid errors in the symbols to direct the x-axis in the direction of elongation of the spring and draw load in the state in which x> 0; spring is stretched. When the calculations can take g = 10 m / s2. Massa springs and connecting rod 2 neglected. The table indicated: c1, c2, c3 - the spring constant, L0 - spring extension with equivalent rigidity at the initial time t = 0, v0 - initial load velocity with respect to the elevator (directed vertically upwards). A dash in columns c1, c2, c3 means that the corresponding spring is not depicted in the drawing and it should not. If the end of one of the other springs will be free, it must be attached in a suitable position, or the load or to the ceiling (the floor) of the elevator; the same should be done if the free bar will be connected to two ends of the two remaining springs. The condition u = 0 means that the strength of the resistance R is absent.

Additional information

After payment you can download and print the file to the solution of the problem in theoretical mechanics D2 Option 15 (which corresponds to Figure 1 and line conditions 5) of manuals for Reshebnik Targ SM 1989 for part-time students of engineering, construction, transport, instrument-specialties of higher education. The task is made in the format word, packed to the archive zip.

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